\newproblem{lay:7_2_24}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.2.24}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Consider the quadratic form $Q(\mathbf{x})=\mathbf{x}^TA\mathbf{x}$ when $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $\det\{A\}\neq 0$. 
	Verify the following statements:
	\begin{enumerate}[a.]
		\item $Q$ is positive definite if $\det\{A\}>0$ and $a+d>0$.
		\item $Q$ is negative definite if $\det\{A\}>0$ and $a+d<0$.
		\item $Q$ is indefinite if $\det\{A\}<0$.
	\end{enumerate}
}{
   % Solution
	\begin{enumerate}[a.]
		\item By definition, $Q$ is positive definite if all its eigenvalues are positive. According to Exercise 7.2.23, $\det\{A\}=\lambda_1\lambda_2$. Then
		      $\det\{A\}=\lambda_1\lambda_2>0$ implies that either both eigenvalues are positive or both are negative. If $a+d>0$, then according to Exercise 7.2.23 $\lambda_1+\lambda_2>0$,
					and both eigenvalues must be positive.
		\item By definition, $Q$ is negative definite if all its eigenvalues are negative. As in the previous point, $\det\{A\}=\lambda_1\lambda_2>0$ implies that
				  either both eigenvalues are positive or both are negative. However, in this case, since $a+d<0$, then $\lambda_1+\lambda_2<0$ and both eigenvalues are negative.
		\item By definition, $Q$ is indefinite if it has positive and negative eigenvalues. If $\det\{A\}=\lambda_1\lambda_2<0$, then both eigenvalues have different sign,
					and consequently $Q$ is indefinite. 
	\end{enumerate}
}
\useproblem{lay:7_2_24}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

